How to Simplify : sin x / ( 1 + cos x ) I want to simplify this expression to contain 2 or fewer terms.  Is this possible, and if so, what are the steps I can take to do so?

sin x / ( 1 + cos x )

 A: I have another idea $1+\cos x=2\cos^2 {\frac{x}{2}}$ and $\sin x=2\sin{\frac{x}{2}}\cos{\frac{x}{2}}$.
Now, the given can be written as $\tan {\frac{x}{2}}$.
A: $$\frac{\sin(x)}{1+\cos(x)}=\frac{\sin(x)}{1+\cos(x)}\frac{1-\cos(x)}{1-\cos(x)}=\frac{\sin(x)(1-\cos(x))}{1-\cos^2(x)}$$
Recall the pythagorean identity to see that $1-\cos^2(x)=\sin^2(x)$.
$$\frac{\sin(x)(1-\cos(x))}{1-\cos^2(x)}=\frac{\sin(x)(1-\cos(x))}{\sin^2(x)}=\frac{1-\cos(x)}{\sin(x)}$$
$$=\frac1{\sin(x)}-\frac{\cos(x)}{\sin(x)}$$

$$=\csc(x)-\cot(x)$$

A: Hint The appearance of $\color{red}{1 + \cos x}$ suggests we can produce an expression without a constant term in the denominator by substituting $x = 2t$ and using the half-angle identity $\cos^2 t = \frac{1}{2}(\color{red}{1 + \cos 2 t})$.

Substituting in our expression gives $$\frac{\sin x}{1 + \cos x} = \frac{\sin 2t}{1 + \cos 2 t} = \frac{2 \sin t \cos t}{2 \cos^2 t} = \frac{\sin t}{\cos t} = \tan t = \tan \frac{x}{2} .$$

A: Use the formulae:
sin x=        (2 tan x/2)/
                      (1+(tanx/2)^2)
cos x=        ( 1-(tan x/2)^2)/
                     (1+(tan x/2)^2)
